Description
Exactly 13.1 ml of water at 27 degrees celsius?
is added to a hot iron skillet. all of the water is converted into steam at 100.0 degrees celsius. the mass of the pan is 1.10 kg and the molar heat capacity of iron is 25.19 J/(mole * degrees celsius). What is the temperature change of the skillet?
Explanation & Answer
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We have to first of all determine the heat required to convert the water at 27 degree Celsius to water at 100 degree Celsius plus the heat required to convert all the liquid at 100 degree C to steam at 100 degree Celsius.
Step 1
Heat required to convert the water from 27 degree Celsius to 100 degree Celsius
=mcdT
m=mass of water =13.1ml=13.1 gram as density is 1g/ml
c=specific heat of water =4.184 J/g degree C
Heat ,H1=13.1*4.184*(100-27)J=4001.1592J=4KJ
Step 2
Calculating the heat required to convert the water at 100 degree C to steam at 100 degree C
Enthalpy of vaporization of water=40.7 KJ/mol
Heat used=40.7*.727 =29.62KJ
Total Heat required by the water=29.62+4 KJ=33.62KJ
Now the heat released by iron=nCDT
n=mole =1100/55.8=19.71 mole
dT=change in temperature
33590.0592=19.71*25.91*dT
dT=65.77 degree C(answer)
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